A metal, M , of atomic mass 56 amu reacts with chlorine to form a salt that can be represented as MClx. A boiling point elevation experiment is performed to determine the subscript x , and therefore, the formula of the salt. A 22.9 g sample of the salt is dissolved in 100.0 g of water and the boiling point of the solution is found to be 375.93 K. Find the formula of the salt. Assume complete dissociation of the salt in solution g

Formula for the salt: MCl₃

Explanation:

MClₓ → M⁺  +  xCl⁻

We apply the colligative property of boiliing point elevation.

We convert the boiling T° to °C

375.93 K - 273K = 102.93°C

ΔT = Kb . m . i

where ΔT means the difference of temperature, Keb, the ebulloscopic constant for water, m the molality of solution (mol of solute/kg of solvent) and i, the Van't Hoff factor (numbers of ions dissolved)

ΔT = 102.93°C - 100°C = 2.93°C

Kb = 0.512 °C/m

We replace data: 2.93°C = 0.512 °C/m . m . i

i = x + 1 (according to the equation)

22.9 g / (56g/m + 35.45x) = moles of salt / 0.1kg = molality

We have calculated the moles of salt in order to determine the molar mass, cause we do not have the data. We replace

2.93°C = 0.512 °C/m . [22.9 g / (56g/m + 35.45x)] / 0.1kg . (x+1)

2.93°C / 0.512 m/°C = [22.9 g / (56g/m + 35.45x)] / 0.1kg . (x+1)

5.72 m = [22.9 g / (56g/m + 35.45x)]/ 0.1 (x+1)

5.72 . 0.1 / [22.9 g / (56g/m + 35.45x)] = x+1

0.572 / (22.9 g / (56g/m + 35.45x) = x+1

0.572 (56 + 35.45x) / 22.9 = x+1

0.572 (56 + 35.45x) = 22.9x + 22.9

32.03 + 20.27x = 22.9x + 22.9

9.13 = 2.62x

x = 3.48 ≅ 3

aquifers are bounded above and below by impermeable rock. the driving force of groundwater movement in confined aquifers is pressure, rather than gravity.